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“问题:确定下列函数的周期: 1) f_1(x)=cos 3x/2-sin x/3 2) f_2(x)=cos 2x-tgx。解:用P表示函数f_1(x)的周期,那末根据周期函数的定义有: cos 3x/2-sin x/3==cos 3/2(x+P)-sin 1/3(x+P)……(1)等式(1)对任何x值都成立。当x=0,就得到: 1-0=cos 3/2P-sin P/3……(2)可知当P=12π时,适合等式(2)。所以函数f_1(x)的周期为12π。类似地可求出f_2(x)的周期。”对于这样的解答,不能使我们满意。第一。“猜测”方程(2)的最小正数解和求出函数f_1(x)最小正周期是同样困难的(或容易的)。因此求出方程(2)来,不能使解答容易。
“Problem: Determine the period of the following function: 1) f_1 (x) = cos 3x/2 - sin x/3 2) f_2 (x) = cos 2x - tgx Solution: Let P denote the period of the function f_1 (x), Then according to the definition of the periodic function: cos 3x/2-sin x/3==cos 3/2(x+P)-sin 1/3(x+P)...(1) Equation (1) for any The value of x holds: when x=0, we get: 1-0=cos 3/2P-sin P/3 (2) It can be seen that when P=12π, it fits equation (2), so the function f_1(x) The period of ) is 12π. Similarly, the cycle of f_2(x) can be found.” For such an answer, we cannot satisfy us. the first. It is equally difficult (or easy) to ”guess" the smallest positive solution of equation (2) and find the smallest positive period of function f_1(x). Therefore, finding equation (2) does not make the solution easy.